Practice Questions
Concept Explanation
Concentration measures the amount of solute in a solution. When mixing solutions, the total amount of solute is conserved while volume changes.
Key Formulas
- Concentration = (amount of solute / total volume) × 100%
- When mixing: C1V1 + C2V2 = Cm(V1 + V2)
- Amount of solute = (concentration/100) × volume
Example
Mix 200 ml of 10% acid with 300 ml of 20% acid. Solute = 20 + 60 = 80 ml in 500 ml = 16%.
Common Trap
Add solute amounts directly, not concentrations. Always divide by total volume for the final concentration.
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Q1 Easy200 ml of 10% acid is mixed with 300 ml of 20% acid. Final concentration is:ExplanationSolute = 0.10 × 200 + 0.20 × 300 = 20 + 60 = 80 ml. Total volume = 500 ml. Concentration = 80 / 500 × 100 = 16%. Hence the answer is 16%.
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Q2 EasyMix 50 L of 12% salt solution with 50 L of 28% salt solution. Final concentration is:ExplanationEqual volumes: mean concentration = (12 + 28) / 2 = 20%. Solute check: 6 + 14 = 20 L in 100 L = 20%. Hence the answer is 20%.
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Q3 EasyPure milk 50 L is mixed with 10 L water. Concentration of milk is:ExplanationMilk = 50 L in total 60 L. Concentration = 50 / 60 × 100 = 250 / 3 = 83 1/3%. Hence the answer is 83 1/3%.
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Q4 MediumA solution of sugar syrup has 15% sugar. How many kg of this syrup should be mixed with 45 kg of a 25% sugar syrup so that the mixture is 20% sugar?ExplanationLet x kg of 15% syrup be mixed. Sugar balance: 0.15x + 0.25 × 45 = 0.20(x + 45). So 0.15x + 11.25 = 0.20x + 9 ⇒ 2.25 = 0.05x ⇒ x = 45 kg. By alligation: (25 − 20) : (20 − 15) = 1 : 1, so x = 45 kg. Hence the answer is 45 kg.
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Q5 MediumA chemist has two solutions of acid. The first is 20% acid and the second is 50% acid. How many litres of each are needed to make 30 litres of 30% acid?ExplanationAlligation for 20% and 50% to get 30%: ratio of 20% : 50% = (50 − 30) : (30 − 20) = 20 : 10 = 2 : 1. In 30 L: 20% solution = (2/3) × 30 = 20 L and 50% solution = 10 L. Check: acid = 0.2 × 20 + 0.5 × 10 = 4 + 5 = 9 L in 30 L = 30%. Hence the answer is 20 L of 20% and 10 L of 50%.
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Q6 MediumHow much water must be added to 40 litres of a 25% acid solution to make it 20% acid?ExplanationAcid amount = 0.25 × 40 = 10 L (unchanged). Let water added = x. Then 10 / (40 + x) = 0.20 ⇒ 10 = 8 + 0.2x ⇒ 0.2x = 2 ⇒ x = 10 L. Hence the answer is 10 L.
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Q7 Medium80 L of 40% alcohol is mixed with 20 L of water. New alcohol concentration is:ExplanationAlcohol = 0.40 × 80 = 32 L. Total volume = 100 L. Concentration = 32%. Hence the answer is 32%.
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Q8 Medium150 ml of 8% solution and 350 ml of 24% solution. Final concentration is:ExplanationSolute = 0.08 × 150 + 0.24 × 350 = 12 + 84 = 96 ml. Total = 500 ml. Concentration = 96 / 500 × 100 = 19.2%. Hence the answer is 19.2%.
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Q9 MediumTwo solutions of 15% and 35% alcohol are mixed in the ratio 1 : 3. Final concentration is:ExplanationMean = (1 × 15 + 3 × 35) / 4 = (15 + 105) / 4 = 120 / 4 = 30%. Hence the answer is 30%.
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Q10 HardAn alloy contains copper and zinc in the ratio 5 : 3. Another alloy has copper : zinc = 7 : 5. Equal quantities of the two alloys are mixed. The ratio of copper to zinc in the mixture is:ExplanationTake 24 units of each alloy (LCM of 8 and 12). First alloy (5 : 3): copper = 15, zinc = 9. Second alloy (7 : 5): copper = 14, zinc = 10. Total copper = 29, total zinc = 19. Ratio copper : zinc = 29 : 19. Hence the answer is 29 : 19.
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Q11 HardAcid and water in a vessel are in the ratio 3 : 2. 20 litres are drawn and replaced by water, after which the ratio becomes 1 : 1. The capacity of the vessel is:ExplanationLet capacity = V. Acid = 3V/5, water = 2V/5. In 20 L drawn: acid = 12 L, water = 8 L. After replacement: acid left = 3V/5 − 12; water = 2V/5 − 8 + 20 = 2V/5 + 12. Set equal for 1 : 1: 3V/5 − 12 = 2V/5 + 12 ⇒ V/5 = 24 ⇒ V = 120 L. Hence the answer is 120 L.
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Q12 HardHow many litres of 50% acid must be mixed with 40 L of 20% acid to get 30% acid?ExplanationAlligation: quantity of 20% : quantity of 50% = (50 − 30) : (30 − 20) = 20 : 10 = 2 : 1. With 40 L of 20%, volume of 50% = 20 L. Check: acid = 0.20 × 40 + 0.50 × 20 = 8 + 10 = 18 L in 60 L = 30%. Hence the answer is 20 L.
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Q13 HardA vessel has 60 L of 30% acid. 12 L is removed and replaced by water. New concentration is:ExplanationAcid removed = 0.30 × 12 = 3.6 L. Acid left = 18 − 3.6 = 14.4 L in 60 L. Concentration = 14.4 / 60 × 100 = 24%. Hence the answer is 24%.
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Q14 HardWhich mixture gives exactly 18% concentration?ExplanationOption A: solute = 10 + 26 = 36 ml in 200 ml = 18%. Option B: 24 + 24 = 48 in 300 = 16%. Option C: 15 + 15 = 30 in 200 = 15%. Option D: 4 + 30 = 34 in 200 = 17%. Hence the answer is A.
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DS1 Easy
A chemist mock asks for litres of the 20% acid solution used. How many litres of the 20% solution are used?
Statement I20% and 50% acids are mixed to make 30 litres of 30% acid.
Statement IIOnly the target concentration 30% is known, not the total volume.
ExplanationStatement I alone: alligation ratio 20% : 50% = 2 : 1, so 20 L of 20% acid in 30 L. Statement II alone is insufficient. Statement I alone is sufficient. Answer: A. -
DS2 Medium
In a lab aptitude item, find how much 15% syrup to mix. How many kg of 15% syrup are needed?
Statement IIt is mixed with 45 kg of 25% sugar syrup to get 20% sugar.
Statement IIOnly the target concentration 20% is known, not either quantity.
ExplanationStatement I alone: solve 0.15x + 0.25 × 45 = 0.20(x + 45) ⇒ x = 45 kg. Statement II alone is insufficient. Statement I alone is sufficient. Answer: A.
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DS3 Medium
Is the final copper : zinc ratio 29 : 19?
Statement ITwo alloys Cu : Zn = 5 : 3 and 7 : 5 are mixed in equal quantities.
Statement IIEqual masses of the two alloys Cu : Zn = 5 : 3 and 7 : 5 are combined.
ExplanationEach statement describes equal mixing of the 5 : 3 and 7 : 5 alloys, giving copper : zinc = 29 : 19. Each alone is sufficient. Answer: D. -
DS4 Hard
Find the capacity of a vessel after replacement. What is the capacity?
Statement IAcid : water was 3 : 2; 20 L drawn and replaced by water makes the ratio 1 : 1.
Statement IIOnly the final ratio 1 : 1 is known, not the initial ratio or the volume drawn.
ExplanationStatement I alone yields V = 120 L from equating acid and water after replacement. Statement II alone is insufficient. Statement I alone is sufficient. Answer: A.